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Convergence to Scattering States in the Nonlinear
Schrödinger Equation
Pascal Bégout
To cite this version:
Pascal Bégout. Convergence to Scattering States in the Nonlinear Schrödinger Equation.
Com-munications in Contemporary Mathematics, World Scientific Publishing, 2001, 3 (3), pp.403-418.
�10.1142/S0219199701000421�. �hal-00715801�
Convergence to Scattering
States in the Nonlinear
Schr¨
odinger Equation
Pascal B´
egout
Laboratoire d’Analyse Num´erique Universit´e Pierre et Marie Curie
Boˆıte Courrier 187
4, place Jussieu 75252 Paris Cedex 05, FRANCE
e-mail : begout@ann.jussieu.fr
Abstract
In this paper, we consider global solutions of the following nonlinear Schr¨odinger equation iut + ∆u + λ|u| α u = 0, in RN , with λ ∈ R, α ∈ (0, 4 N−2) (α ∈ (0, ∞) if N = 1) and u(0) ∈ X ≡ H1(RN
) ∩ L2(|x|2; dx). We show that, under suitable conditions, if the solution
usatisfies e−it∆u(t) − u±→ 0 in X as t → ±∞ then u(t) − eit∆
u±→ 0 in X as t → ±∞. We
also study the converse. Finally, we estimate | ku(t)kX− ke it∆
u±kX| under some less restrictive
assumptions.
1
Introduction and notations
We consider the following Cauchy problem, i∂u ∂t + ∆u + λ|u| α u = 0, (t, x) ∈ (−T∗, T∗) × RN, u(0) = ϕ, in RN, (1.1) where λ ∈ R, 0 6 α < 4
N − 2 (0 6 α < ∞ if N = 1) and ϕ a given initial data. It is well-known that if λ < 0, α > 4
N and ϕ ∈ H
1(RN), then there exists u
±∈ H1(RN) such that
lim
t→±∞kT (−t)u(t)−u±kH1 = 0 (Ginibre and Velo [8], Nakanishi [11,12]). Since (e it∆)
t∈Ris an isometry
on H1(RN), we also have lim
t→±∞ku(t) − T (t)u±kH1= 0. Furthermore, if α >
−(N−2)+√N2+12N +4
2N and
if ϕ ∈ X ≡ H1(RN) ∩ L2(|x|2; dx), then there exist u
±∈ X such that limt
→±∞kT (−t)u(t) − u±kX = 0
(Tsutsumi [15]). The same result holds without assumption on the λ’s sign if the initial data is small
enough in X and if α > 4
N + 2 (Cazenave and Weissler [3]). Note that to have these limits, we have to make a necessary assumption on α (Barab [1], Strauss [13, 14], Tsutsumi and Yajima [16]),
2 N < α <
4
N − 2 (2 < α < ∞ if N = 1).
The purpose of this paper is to study the asymptotic behavior of ku(t) − T (t)u±kX under the
assumption lim
t→±∞kT (−t)u(t) − u±kX = 0, and the converse. In the linear case (i.e. : λ = 0) or if the
initial data is 0, the answer is trivial since T (−t)u(t) − u± ≡ u(t) − T (t)u±≡ 0, for all t ∈ R. Since
(eit∆)
t∈R is an isometry on H1(RN), the equivalence on H1(RN) is trivial. But (eit∆)t∈R is not an
isometry on X and so it is natural to wonder whether or not we have lim
t→±∞ku(t) − T (t)u±kX = 0
when lim
t→±∞kT (−t)u(t) − u±kX = 0 and conversely.
This paper is organized as follows. In Section2, we give the main results. In Section3, we establish some a priori estimates. In Section4, we prove Theorems2.1,2.4,2.5and Proposition2.8. In Section 5, we prove Theorem2.10.
Before closing this section, we give some notations which will be used throughout this paper and we recall some properties of the solutions of the nonlinear Schr¨odinger equation.
z is the conjugate of the complex number z; Re z and Im z are respectively the real and imaginary part of the complex number z; ∆ = PN
j=1 ∂2
∂x2
j; for 1 6 p 6 ∞, p
′is the conjugate of the real number p
de-fined by1p+p1′ = 1 and Lp= Lp(RN) = Lp(RN; C) with norm k.kLp; H1= H1(RN) = H1(RN; C) with
norm k . kH1; for all (f, g) ∈ L2× L2, (f, g) = Re R
RN f (x)g(x)dx; X =ψ ∈ H1(RN; C); kψk X< ∞ with norm kψk2 X = kψk2H1(RN)+ R RN |x|2|ψ(x)|2dx; (T (t))
t∈Ris the group of isometries (eit∆)t∈R
gen-erated by i∆ on L2(RN; C); C are auxiliary positive constants and C(a
1, a2, . . . , an) indicates that
the constant C depends only on parameters a1, a2, . . . , an and that the dependence is continuous.
It is well-known that for every ϕ ∈ X, (1.1) has a unique solution u ∈ C((−T∗, T∗); X) which
satisfies the conservation of charge and energy, that is for all t ∈ (−T∗, T∗), ku(t)kL2 = kϕkL2 and
E(u(t)) = E(ϕ)def= 1 2k∇ϕk
2
L2−α+2λ kϕk
α+2
Lα+2. Moreover, if λ 6 0, if α < N4 or if kϕkH1 is small enough
then T∗= T
∗= ∞ and kukL∞(R;H1)< ∞ (see for example Cazenave [2], Ginibre and Velo [4,5,6,7],
Kato [9]).
Definition 1.1. We say that (q, r) is an admissible pair if the following holds. (i) 2 6 r 6 N−22N (2 6 r < ∞ if N = 2, 2 6 r 6 ∞ if N = 1),
(ii) 2q = N 12−1r .
Note that in this case 2 6 q 6 ∞ and q = 4r N (r − 2).
Definition 1.2. We say that a solution u ∈ C((−T∗, T∗); X) of (1.1) has a scattering state u+ at
+∞ (respectively u− at −∞) if T∗ = ∞ and if u+ ∈ X is such that lim
t→∞kT (−t)u(t) − u+kX = 0
(respectively if T∗= ∞ and if u−∈ X is such that limt
→−∞kT (−t)u(t) − u−kX= 0).
We recall the Strichartz’ estimates. Let I ⊆ R, be an interval, let t0 ∈ I, let (q, r) and (γ, ρ)
be two admissible pairs, let ϕ ∈ L2(RN) and let f ∈ Lγ′
(I; Lρ′
(RN)). Then the following integral
equation defined for all t ∈ I, u(t) = T (t)ϕ + i Z t
t0
T (t − s)f(s)ds, satisfies the following inequality kukLq(I,Lr)6C0kϕkL2+C1kfkLγ′(I;Lρ′), where C0= C0(N, r) and C1= C1(N, r, ρ). For more details,
see Keel and Tao [10].
2
The main results
Theorem 2.1. Let λ 6= 0, 2
N < α < 4
N − 2 (2 < α < ∞ if N = 1), ϕ ∈ X and let u be the solution of (1.1) such that u(0) = ϕ. We assume that u has a scattering state u± at ±∞ (see Definition 1.2). Then the following holds.
1. (a) If N 6 2 and if α > 4
N then limt→±∞ku(t) − T (t)u±kX= 0.
(b) If 3 6 N 6 5 and if α > 8
N + 2 then limt→±∞ku(t) − T (t)u±kX= 0.
2. If N = 1 and α = 4 or if 3 6 N 6 5 and α = 8
N + 2 then we have, sup
t>0ku(t) − T (t)u+kX < ∞ and
sup
t60ku(t) − T (t)u−kX< ∞.
Remark 2.2. Remark that in Theorem2.1, no hypothesis on the λ′ s sign is made.
Remark 2.3. N ∈ {3, 4, 5} =⇒ 4 N < 8 N + 2 < 6 N < 4 N − 2. N > 6 =⇒ 4 N − 2 6 8 N + 2.
Despite the fact we do not know if lim
t→±∞ku(t) − T (t)u±kX = 0 when α 6
8
N + 2 (α 6 4/N if N 6 2) or when N > 6, we can give an estimate of the difference of the norms, as shows the following theorem, without any restriction on the dimension space N and on α (except α > 2
N). Since under the
scattering state assumption we always have lim
t→±∞ku(t) − T (t)u±kH1 = 0, it is sufficient to estimate
Theorem 2.4. Let λ < 0, 2 N < α <
4
N − 2 (2 < α < ∞ if N = 1), ϕ ∈ X and let u be the associated solution of (1.1). Assume that u has a scattering state u± at ±∞ (see Definition 1.2). We define for all t ∈ R, A±(t) = kxu(t)kL2− kxT (t)u±kL2 and h(t) = kxu(t)k2
L2. Then we have the following result.
sup
t>0| ku(t)kX− kT (t)u+kX| < ∞ and
sup
t60| ku(t)kX− kT (t)u−kX| < ∞,
with the following estimates. 1. If α < 4 N then − C k∇u±kL2 6lim inf t→±∞A±(t) 6 lim supt→±∞ A±(t) 6 ± h′(0) + 4(xu ±, i∇u±) 4k∇u±kL2 . 2. If α > 4 N then ± h′(0) + 4(xu±, i∇u±) 4k∇u±kL2 6lim inf t→±∞A±(t) 6 lim supt→±∞ A±(t) 6 C k∇u±kL2. 3. If α = 4 N then limt→±∞A±(t) = ± h′(0) + 4(xu ±, i∇u±) 4k∇u±kL2 . Furthermore, h′(0) = 4Im Z RN ϕ(x)x.∇ϕ(x)dx and C = C(sup t∈RkT (−t)u(t)kX , N, α, λ). Theorem 2.5. Let λ > 0, 2 N < α < 4
N − 2 (2 < α < ∞ if N = 1), ϕ ∈ X and let u be the associated solution of (1.1). Assume that u has a scattering state u± at ±∞ (see Definition 1.2). We define for
all t ∈ R, A±(t) = kxu(t)kL2− kxT (t)u±kL2 and h(t) = kxu(t)k2
L2. Then we have the following result.
sup
t>0| ku(t)kX− kT (t)u+kX| < ∞ and
sup
t60| ku(t)kX− kT (t)u−kX| < ∞,
with the following estimates. 1. If α < 4 N then ± h′(0) + 4(xu ±, i∇u±) 4k∇u±kL2 6lim inf t→±∞A±(t) 6 lim supt→±∞ A±(t) 6 C k∇u±kL2 . 2. If α > 4 N then − C k∇u±kL2 6lim inf t→±∞A±(t) 6 lim supt→±∞ A±(t) 6 ± h′(0) + 4(xu ±, i∇u±) 4k∇u±kL2 . 3. If α = 4 N then limt→±∞A±(t) = ± h′(0) + 4(xu ±, i∇u±) 4k∇u±kL2 . Furthermore, h′(0) = 4Im Z RN ϕ(x)x.∇ϕ(x)dx and C = C(sup t∈RkT (−t)u(t)k X, N, α, λ).
Remark 2.6. By Theorem2.1and2of Theorems2.4and2.5, if α > 4
N when N 6 2 or if α > 8 N + 2 when N ∈ {3, 4, 5}, we have Im Z RN u−(x)x.∇u−(x)dx 6 Im Z RN ϕ(x)x.∇ϕ(x)dx 6 Im Z RN u+(x)x.∇u+(x)dx, if λ < 0, Im Z RN u+(x)x.∇u+(x)dx 6 Im Z RN ϕ(x)x.∇ϕ(x)dx 6 Im Z RN u−(x)x.∇u−(x)dx, if λ > 0.
Remark 2.7. If ϕ ∈ H1(RN; C) satisfies ϕ ≡ aψ with ψ ∈ H1(RN; R) and a ∈ C, then we have h′(0) ≡ d dtkxu(t)k 2 L2|t=0≡ 4Im Z RN ϕ(x)x.∇ϕ(x)dx = 0.
The following proposition offers others estimates. Proposition 2.8. Let λ 6= 0, 2
N < α < 4
N − 2 (2 < α < ∞ if N = 1), ϕ ∈ X and let u be the solution of (1.1) such that u(0) = ϕ. Assume that u has a scattering state u± at ±∞. Then the
following estimates hold. 1. If λ < 0, lim sup
t→±∞(kxu(t)kL
2− kxT (t)u±kL2) 6 kxu±kL
2k∇u±kL2± (xu±, i∇u±)
k∇u±kL2
.
2. If λ > 0, lim inf
t→±∞(kxu(t)kL2− kxT (t)u±kL2) > −
kxu±kL2k∇u±kL2∓ (xu±, i∇u+)
k∇u±kL2
.
Remark 2.9. By3of Theorems2.4 and2.5and by Proposition2.8, if α = 4
N then we have, −kxu−kL2k∇u−kL2 61 4 d dtkxu(t)k 2 L2|t=06kxu+kL2k∇u+kL2, if λ < 0, −kxu+kL2k∇u+kL26 1 4 d dtkxu(t)k 2 L2|t=06kxu−kL2k∇u−kL2, if λ > 0.
Now we give the result concerning the converse. Theorem 2.10. Let λ 6= 0, 2
N < α < 4
N − 2 (2 < α < ∞ if N = 1), ϕ ∈ X and u be the associated solution of (1.1). Assume that u is global in time and there exists u+ ∈ X and u− ∈ X such that
lim
t→±∞ku(t) − T (t)u±kX = 0. Let α0 =
−(N − 2) +√N2+ 12N + 4
2N . Then, we have the following result.
1. If λ < 0 and if α > α0 (α > α0 if N = 2) then lim
t→±∞kT (−t)u(t) − u±kX = 0.
2. If λ > 0 and if α > 4
N then limt→±∞kT (−t)u(t) − u±kX= 0.
3. If α > 4
N + 2 and if kϕkX is small enough then limt→±∞kT (−t)u(t) − u±kX= 0.
Remark 2.11. Note that in the case2, no hypothesis on the kϕkX’ s size is made.
Remark 2.12. Assume there exists u±, v± ∈ X such that limt
→±∞kT (−t)u(t) − u±kX = 0 and
lim
t→±∞ku(t) − T (t)v±kX = 0. Then we have, u+= v+ and u−= v−. Indeed, since X ֒→ L
2(RN) and
T (t) is an isometry on L2(RN), we have lim
t→±∞kT (−t)u(t) − u±kL2= limt→±∞kT (−t)u(t) − v±kL2= 0.
Hence the result. Remark 2.13. α0∈ 4 N+2, 4 N (α0∈ N2,N4 if N = 1).
3
A priori estimates
Throughout this section, we make the following assumptions.
λ 6= 0, N2 < α < 4
N − 2 (2 < α < ∞) if N = 1), ϕ ∈ X, u ∈ C(R; X) is the associated solution of (1.1) and has a scattering state u±∈ X at ± ∞.
(3.1)
We define the following real.
γ∗= α − 2 2 , if N = 1, α(N + 2) − 4 4 , if N > 2. (3.2)
Proposition 3.1. Assume u satisfies (3.1) (we can suppose instead of u has a scattering state that we only have sup
t∈RkT (−t)u(t)kX< ∞). Let (q, r) be an admissible pair (see Definition
1.1). Then the following holds.
1. For all t 6= 0, ku(t)kLr 6C|t|− 2 q, where C = C(sup t∈RkT (−t)u(t)kX , N, r). 2. If furthermore α > 4 N + 2 then u ∈ L q(R; W1,r(RN)).
Proof. We follow the method of Cazenave [2] (see Theorem 7.2.1 and Corollary 7.2.4). We set w(t, x) = e−i|x|24t u(t, x) and f (u) = λ|u|αu. We already know that for every admissible pair (q, r),
u ∈ Lqloc(R; W1,r(RN)) (see for example Cazenave [2]; Theorem 5.3.1 and Remark 5.3). We only
prove the case t > 0, the case t < 0 following by applying the result for t > 0 to u(−t) solution of (1.1) with initial value ϕ. We proceed in 2 steps.
Step 1. ku(t)kLr6C(sup
t∈RkT (−t)u(t)kX, N, r)|t| −2
q, for every admissible pair (q, r) and for all t 6= 0.
We have kxT (−t)u(t)kL2= k(x+2it∇)u(t)kL2 6C. Furthermore, (x+2it∇)u(t, x) = 2ite−i |x|2
4t ∇w(t, x).
Using the Gagliardo-Nirenberg’s inequality, we obtain
ku(t)kLr ≡ kw(t)kLr 6Ck∇w(t)k N(1 2− 1 r) L2 kw(t)k 1−N(1 2− 1 r) L2 6C k(x + 2it∇)u(t)kL2|t|−1 N(1 2− 1 r) 6C|t|−N(12− 1 r).
Hence the result.
By the Strichartz’ estimates and by H¨older’s inequality (applying twice), we have kf(u)kLq′((0,∞);W1,r′)6C kukLqα q−2((0,1);Lr−2rα )+ kukLq−2qα ((1,∞);Lr−2rα ) α kukLq((0,∞);W1,r), (3.3) kukLq((S,∞);W1,r)6C + Ckukα L qα q−2((S,∞);Lr−2rα )kukLq((S,∞);W 1,r), (3.4)
for all S > 0 and for every admissible pair (q, r). Case N >3. We set r = 4N 2N −α(N−2). Since α ∈ 0, 4 N−2 , we have r ∈2, 2N N−2 . So we can take q such that (q, r) is an admissible pair. For this choice of r, we have rα
r−2 = 2N N−2 and q q−2 = 4 4−α(N−2).
By (3.4) and the first step we have for all S > 0,
kukLq((S,∞);W1,r)6C + C Z ∞ S t−4−α(N −2)4α dt q−2q kukLq((S,∞);W1,r). And 4α 4 − α(N − 2) > 1 ⇐⇒ α > 4
N + 2. Thus, there exists S0> 0 large enough such that
C Z ∞ S0 t−4−α(N −2)4α dt q−2q 6 1 2, and then, kukLq((S0,∞);W1,r)62C.
For this choice of (q, r), we deduce from (3.3) that kf(u)kLq′((0,∞);W1,r′)< ∞. Hence the result for
every admissible pair by the Strichartz’ estimates.
Case N=2. Since α > 1 is fixed, we take r > 2 sufficiently close to 2 to have α > 2(r − 1) r . So, in particular, rα r − 2 > 2. Moreover, q q − 2 = r
2 where q is such that (q, r) is an admissible pair. So by H¨older’s inequality (twice), (3.4) and the first step, we have for all S > 0,
kukLq((S,∞);W1,r)6C + C Z ∞ S t−rα−2(r−2)2 dt 2r kukLq((S,∞);W1,r). And rα − 2(r − 2) 2 > 1 ⇐⇒ α > 2(r − 1)
r . And we conclude exactly as the case N > 3. Case N=1. We take (3.3) with the admissible pair (∞, 2) and apply the first step. So,
kf(u)kL1((0,∞);H1)6C kukLα((0,1);L∞)+ kukLα((1,∞);L∞)
α kukL∞((0,∞);H1) 6C + CkukαLα((1,∞),L∞)6C + C ∞ Z 1 t−α2dt < ∞.
Remark 3.2. We set v(t, x) = (x + 2it∇)u(t, x). In the same way as the above proof, we can show under the assumptions of Proposition 3.1 that if α > 4
N + 2, then for every admissible pair (q, r), v ∈ Lq(R; Lr(RN)) (follow the step 2 of the proof of Corollary 7.2.4 of Cazenave [2]; consider separately
the three cases N = 1, N = 2, N > 3 and replace the admissible pairs therein by those of the proof of Proposition3.1).
Proposition 3.3. Let γ∗ be defined by (3.2). Assume u satisfies (3.1) and α > 4
N + 2. Then, the following estimates hold.
1. If N = 1 then for all t 6= 0 we have kT (−t)u(t) − u±kH1 6C|t|−γ ∗
.
2. If N = 2 then for all t 6= 0 and for any γ < γ∗, we have kT (−t)u(t) − u
±kH1 6C|t|−γ.
3. If N > 3 then for all t 6= 0 we have kT (−t)u(t) − u±kH1 6C|t|−γ ∗
.
Proof. Denote f (u) = λ|u|αu. We only prove the case t > 0, the case t < 0 following by applying the result for t > 0 to v(t) = u(−t) solution of (1.1) with v(0) = ϕ. In this case, v+ = u− and the result
follows. By applying the Strichartz’ estimates and H¨older’s inequality (twice), we have
ku(t) − T (t)u+kH16Ckf(u)kLq′(t,∞);W1,r′)6Ckukα
L
qα
q−2((t,∞);Lr−2rα )kukL
q(R;W1,r),
for every admissible pair (q, r) and for all t > 0. Thus, by Proposition3.1,2, we have
ku(t) − T (t)u+kH1 6Ckukα
L
qα
q−2((t,∞);Lr−2rα ),
for every admissible pair (q, r) and for all t > 0.
Now, we conclude by the same way than for the step 2 of the proof of Proposition 3.1, using 1 of this proposition, considering separately the three cases N = 1, N = 2, N > 3, and using the same admissible pairs. This achieves the proof.
4
Proof of Theorems
2.1
,
2.4
,
2.5
and Proposition
2.8
Throughout this section, we assume that u satisfies (3.1).Proof of Theorem 2.1. Since lim
t→±∞kT (−t)u(t) − u±kX = 0 and T (t) is an isometry on H 1, we
have lim
t→±∞ku(t) − T (t)u±kH1 = 0. Thus, it is sufficient to prove that limt→±∞kxu(t) − xT (t)u±kL2= 0
to obtain 1 and that sup
t>0kxu(t) − xT (t)u
+kL2 < ∞ and sup
t60kxu(t) − xT (t)u−kL
2. Suppose that the result is proved for t > 0. Then we apply it to v(t) = u(−t) solution of (1.1) with initial data v(0) = ϕ. Then v+ = u− is the scattering state at +∞ of u(−t). And using the
identity T (t)ψ = T (−t)ψ which holds for all t ∈ R and for every ψ ∈ L2, we obtain the result for t < 0. So to conclude, it is sufficient to prove that lim
t→∞kxu(t) − xT (t)u+kL2 = 0 to obtain 1 and
sup
t>0kxu(t) − xT (t)u+kL
2< ∞ to obtain2. We have
xu(t) − xT (t)u+= xu(t) − T (t)xu++ 2itT (t)∇u+
= xu(t) + 2it∇u(t) − T (t)xu++ 2itT (t)∇u+− 2it∇u(t)
= T (t) [(xT (−t)u(t) − xu+) + 2it(∇u+− T (−t)∇u(t))] ,
for all t > 0. With Proposition3.3, we obtain
kxu(t) − xT (t)u+kL26kxT (−t)u(t) − xu+kL2+ 2tkT (−t)∇u(t) − ∇u+kL2
6kxT (−t)u(t) − xu+kL2+ Ct−(γ−1),
for all t > 0 and for all γ ∈ (0, γ∗] if N 6= 2 and for all γ ∈ (0, γ∗) if N = 2, where γ∗ is defined by (3.2). And by assumption, lim
t→∞kxT (−t)u(t) − xu+kL
2 = 0 and γ∗− 1 > 0 if and only if α > 8
N+2 if
N > 3 and α > 4
N if N 6 2. Thus, in the above expression, it is sufficient to choose γ = γ∗ if N 6= 2,
and γ ∈ (0, γ∗) close enough to γ∗ if N = 2. Hence the result.
Remark 4.1. Since we do not have the estimate 2 of Proposition3.3 for γ = γ∗, we do not know
whether or not sup
t>0ku(t) − T (t)u
+kX< ∞ and sup
t60ku(t) − T (t)u−kX< ∞ when N = 2 and α =
4 N. We define the following function h by
∀t ∈ (−T∗, T∗), h(t) = kxu(t)k2L2. (4.1)
Lemma 4.2. Let u satisfying (3.1) and let h be defined by (4.1). Then h ∈ C2(R), and we have
∀t ∈ R, h′(t) = 4ImZ RN u(t, x)x.∇u(t, x)dx, h′′(t) = 2N αk∇u +k2L2− 2(Nα − 4)k∇u(t)k2L2. Furthermore, if λ < 0, ∀t ∈ R, k∇u(t)kL26k∇u+kL2, (4.2) and if λ > 0, ∀t ∈ R, k∇u(t)kL2>k∇u+kL2. (4.3)
Proof. See Ginibre and Velo [6] or Cazenave [2], Proposition 6.4.2 to have h ∈ C2(R), the expression
of h′ and ∀t ∈ R, h′′(t) = 4N αE(ϕ) − 2(Nα − 4)k∇u(t)k2
L2. Furthermore, using the conservation of
energy and lim
t→±∞ku(t)kLα+2= 0 (Proposition3.1), we obtain k∇u+k 2
L2 = 2E(ϕ). Which gives, with
the above identity, the desired expression of h′′. Finally, with the equality k∇u
+k2L2= 2E(ϕ) and the
conservation of energy, we obtain (4.2) if λ < 0 and (4.3) if λ > 0.
Now, we establish 2 lemmas which will be used to prove Theorem2.4.
Lemma 4.3. Let u satisfying (3.1) and let h be defined by (4.1). Assume that λ < 0. Then the following holds.
(i) If α < 4
N then lim supt→∞ (kxu(t)kL
2− kxT (t)u+kL2) 6 h
′(0) + 4(xu+, i∇u+)
4k∇u+kL2 .
(ii) If α > 4
N then lim inft→∞ (kxu(t)kL2− kxT (t)u+kL2) >
h′(0) + 4(xu
+, i∇u+)
4k∇u+kL2
.
(iii) If α = 4
N then limt→∞(kxu(t)kL2− kxT (t)u+kL2) =
h′(0) + 4(xu
+, i∇u+)
4k∇u+kL2 .
Proof. We proceed in 4 steps. Step 1. (a) If α < 4 N then ∀t > 0, h′(t) > h′(0) + 2N αk∇u+k 2 L2t. (b) If α > 4 N then ∀t > 0, h′(t) > h′(0) + 8k∇u+k 2 L2t. (c) If α = 4 N then ∀t ∈ R, h′(t) = h′(0) + 8k∇u+k 2 L2t.
We integrate between 0 and t > 0 the function h′′ of Lemma4.2.
α 6 4 N =⇒ −2(Nα − 4) > 0 =⇒ (a). α > 4 N =⇒ −2(Nα − 4) 6 0 with (4.2) =⇒ (b). α = 4 N =⇒ −2(Nα − 4) = 0 =⇒(c). Step 2. (a) If α < 4 N then ∀t > 0, h′(t) 6 h′(0) + 8k∇u+k 2 L2t. (b) If α > 4 N then ∀t > 0, h ′(t) 6 h′(0) + 2N αk∇u +k2L2t.
We integrate between 0 and t > 0 the function h′′ of Lemma4.2.
α < 4 N =⇒ −2(Nα − 4) > 0 and (4.2) =⇒(a). α > 4 N =⇒(b). Step 3. (a) If α < 4 N then ∀t > 0, kxϕk2L2+ h′(0)t + N αk∇u+k2L2t26h(t) 6 kxϕk2L2+ h′(0)t + 4k∇u+k2L2t2.
(b) If α > 4 N then ∀t > 0, kxϕk2 L2+ h′(0)t + 4k∇u+k2L2t26h(t) 6 kxϕk2L2+ h′(0)t + N αk∇u+k2L2t2. (c) If α = 4 N then ∀t ∈ R, h(t) = kxϕk 2 L2+ h′(0)t + 4k∇u+k2L2t2.
It is sufficient to integrate between 0 and t > 0 the formulas of steps 1 and 2 to obtain the step 3. Step 4. Conclusion.
We set : g(t) =qkxϕk2
L2+ h′(0)t + 4k∇u+k2L2t2, t > 0 large enough.
Then for t > 0 large enough, we have the following asymptotic development:
g(t) = 2k∇u+kL2t + h ′(0) 4k∇u+kL2 + kxϕk 2 L2 4k∇u+kL2t + o 1 t . (4.4)
In the same way, for t > 0 large enough, we have :
kxT (t)u+kL2= 2k∇u+kL2t − (xu+, i∇u+) k∇u+kL2 + kxu+k 2 L2 4k∇u+kL2t + o 1 t . (4.5)
And, applying the step 3 (a), (4.4) and (4.5) and taking lim sup
t→∞ , we get (i). Indeed,
kxu(t)kL2− kxT (t)u+kL2 6 2k∇u+kL2t + h′(0) 4k∇u+kL2 + kxϕk2 L2 4k∇u+kL2t− 2k∇u+kL 2t + (xu+, i∇u+) k∇u+kL2 − kxu+k2L2 4k∇u+kL2t+ o 1 t = h′(0) 4k∇u+kL2 + kxϕk 2 L2 4k∇u+kL2t +(xu+, i∇u+) k∇u+kL2 − kxu+k2L2 4k∇u+kL2t + o 1 t .
Hence (i) by taking lim sup
t→∞
in the above expression.
By applying the step 3 (b), (4.4) and (4.5) and taking lim inf
t→∞ , we get (ii) by the same way.
By applying the step 3 (c), (4.4) and (4.5) and letting t −→ ∞, we get (iii) by the same way.
Lemma 4.4. Let u satisfying (3.1). Assume that λ < 0. Then the following holds. (i) If α < 4
N then lim inft→∞ (kxu(t)kL2− kxT (t)u+kL2) > −
C k∇u±kL2
.
(ii) If α > 4
N then lim supt→∞ (kxu(t)kL
2− kxT (t)u+kL2) 6 C k∇u±kL2 . Furthermore, C = C(sup t∈RkT (−t)u(t)kX , N, α, λ).
Proof. We proceed in 2 steps. Let h be defined by (4.1). Step 1. ∀t > 1, − t Z 1 s Z 1 k∇u(σ)k2L2dσds 6 C + Ct− N α−4 2 + Ct + k∇u +k2t − 1 2k∇u+k 2 L2t2, where
C = C(sup
t∈RkT (−t)u(t)kX
, N, α, λ).
By Proposition3.1, we have ku(σ)kα+2Lα+2 6C(sup
t∈RkT (−t)u(t)kX
, N, α)σ−N α
2 , for all σ > 0. With the
conservation of energy and the formula k∇u+k2L2= 2E(ϕ), we deduce that for all σ > 0,
k∇u+k2L2− k∇u(σ)k2L26C(sup
t∈RkT (−t)u(t)kX
, N, α, λ)σ−N α2
(since λ < 0). Integrating this expression over [1, t] × [1, s], we obtain the desired result. Step 2. Conclusion.
(i) Lemma4.2implies that for every t > 0,
h(t) = kxϕk2L2+ h′(0)t + N αk∇u+k2L2t2− 2(Nα − 4) t Z 0 s Z 0 k∇u(σ)k2L2dσds. (4.6)
By (4.6) and step 1, we obtain,
∀t > 1, h(t) > C + h′(0)t + N αk∇u+k2L2t2+ Ct 4−N α
2 − Ct + 2(Nα − 4)k∇u+k2
L2t − (Nα − 4)k∇u+k2t2.
And so for all t > 1, h(t) > C + Ct4−N α2 + (h′(0) − 2(4 − Nα)k∇u+k2
L2− C)t + 4k∇u+k2L2t2. By an
asymptotic development on this last inequality, we obtain
kxu(t)kL2>2k∇u+kL2t + Ct−1+ Ct− N α−2 2 +h ′(0) − 2(4 − Nα)k∇u +k2L2− C 4k∇u+kL2 + o 1 t ,
for all t > 0 large enough. From this last expression and (4.5), we obtain (i) (see the end of the proof of (i) of Lemma4.3).
(ii) From (4.6) and step 1, we obtain, for all t > 1,
h(t) 6 C + f′(0)t + N αk∇u+k2L2t2+ Ct 4−N α
2 + Ct + 2(N α − 4)k∇u
+k2L2t − (Nα − 4)k∇u+k2L2t2.
And so, for all t > 1, h(t) 6 C + Ct4−N α2 + (h′(0) + 2(N α − 4)k∇u+k2
L2+ C)t + 4k∇u+k2L2t2. By an
asymptotic development on this last inequality, we obtain
kxu(t)kL262k∇u+kL2t + Ct−1+ Ct− N α−2 2 +h ′(0) − 2(4 − Nα)k∇u+k2 L2+ C 4k∇u+kL2 + o 1 t ,
for all t > 0 large enough. With this last expression and (4.5), we obtain (ii) (see the end of the proof of (i) of Lemma4.3).
Now, we are able to prove Theorem 2.4.
| kxu(t)kL2 − kxT (t)u+kL2| remains bounded as t −→ ∞. Lemmas 4.3 and 4.4 achieve the proof
and give the desired estimates.
Now, we establish 2 lemmas which will be used to prove Theorem 2.5. The proof is very similar to the Theorem2.4one.
Lemma 4.5. Let u satisfying (3.1) and let h be defined by (4.1). Assume that λ > 0. Then the following holds.
(i) If α < 4
N then lim inft→∞ (kxu(t)kL2− kxT (t)u+kL2) >
h′(0) + 4(xu
+, i∇u+)
4k∇u+kL2
.
(ii) If α > 4
N then lim supt→∞ (kxu(t)kL
2− kxT (t)u+kL2) 6 h
′(0) + 4(xu+, i∇u+)
4k∇u+kL2 .
(iii) If α = 4
N then limt→∞(kxu(t)kL2− kxT (t)u+kL2) =
h′(0) + 4(xu+, i∇u+)
4k∇u+kL2
.
Proof. We proceed as for the proof of Lemma4.3, using (4.3) instead of (4.2).
Lemma 4.6. Let u satisfying (3.1). Assume that λ > 0. Then the following holds. (i) If α < 4
N then lim supt→∞ (kxu(t)kL
2− kxT (t)u+kL2) 6 C
k∇u±kL2.
(ii) If α > 4
N then lim inft→∞ (kxu(t)kL2− kxT (t)u+kL2) > −
C k∇u±kL2 . Furthermore, C = C(sup t∈RkT (−t)u(t)kX , N, α, λ).
Proof. We proceed in 2 steps. Step 1. ∀t > 1, − t Z 1 s Z 1 k∇u(σ)k2L2dσds > −C − Ct− N α−4 2 − Ct + k∇u +k2L2t − 1 2k∇u+k 2 L2t2, where C = C(sup t∈RkT (−t)u(t)kX , N, α, λ).
By Proposition3.1, we have ku(σ)kα+2Lα+2 6C(sup
t∈RkT (−t)u(t)kX
, N, α)σ−N α
2 , for all σ > 0. With the
conservation of energy and the equality k∇u+k2L2= 2E(ϕ), we deduce that for all σ > 0,
k∇u+k2L2− k∇u(σ)k2L2 >−C(sup
t∈RkT (−t)u(t)kX
, N, α, λ)σ−N α2
(since λ > 0). Integrating this expression over [1, t] × [1, s], we get the desired result. Step 2. Conclusion.
First, note that we have for all t > 1, Z t 0 Z s 0 k∇u(σ)k 2 L2dσds = Z t 0 Z 1 0 k∇u(σ)k 2 L2dσds + Z 1 0 Z s 1 k∇u(σ)k 2 L2dσds + Z t 1 Z s 1 k∇u(σ)k 2 L2dσds 6 Z t 0 Z 1 0 k∇u(σ)k 2 L2dσds + Z t 1 Z s 1 k∇u(σ)k 2 L2dσds. And so, t Z 0 s Z 0 k∇u(σ)k2L2dσds 6 C(kϕkH1, N, α, λ)t + t Z 1 s Z 1 k∇u(σ)k2L2dσds,
for all t > 1. And we proceed as for the proof of Lemma4.4. Now, we are able to prove Theorem 2.5.
Proof of Theorem 2.5. As for the proof of Theorem 2.1, it is sufficient to show that | kxu(t)kL2 − kxT (t)u+kL2| remains bounded as t −→ ∞. Lemmas 4.5 and 4.6 achieve the proof
and give the desired estimates.
Proof of Proposition 2.8. By Cauchy-Schwarz’ inequality, we have
| kxu(t)kL2− 2tk∇u(t)kL2| 6 kxT (−t)u(t)kL2, (4.7)
for all t ∈ R and for every λ 6= 0. We have also the following estimate.
kxT (t)u+kL2= 2tk∇u+kL2−(xu+, i∇u+)
k∇u+kL2 + kxu+k2L2 4k∇u+kL2t + o 1 t , (4.8)
for all t > 0 large enough and for every λ 6= 0.
We first establish 1 in the case t > 0 that we note in this proof1+. 1 in the case t < 0 is obviously
noted1−. By (4.7), (4.8) and (4.2), we have kxu(t)kL2− kxT (t)u+kL2
6 2tk∇u(t)kL2+ kxT (−t)u(t)kL2− 2tk∇u+kL2+(xu+, i∇u+)
k∇u+kL2 − kxu+k2L2 4k∇u+kL2t+ o 1 t
6 kxT (−t)u(t)kL2+(xu+, i∇u+)
k∇u+kL2 − kxu+k2L2 4k∇u+kL2t+ o 1 t
for all t > 0 large enough. Hence1+ by taking lim sup
such that v(0) = ϕ. Indeed, by uniqueness v(t) = u(−t) and so v+ = u−. Then, using 1+ and the
identity T (t)ψ = T (−t)ψ which holds for all t ∈ R and ψ ∈ L2, we obtain1 −.
Using (4.3) instead of (4.2), we obtain 2by the same way.
5
Proof of Theorem
2.10
Proof of Theorem 2.10. We proceed in 2 steps. Step 1. We have 1and3.
It is well-known that there exist v±∈ X such that T (−t)u(t)−−−−→X
t→±∞ v± (Cazenave and Weissler [3]).
The result follows from Remark2.12. Step 2. We have 2.
We set v(t, x) = (x + 2it∇)u(t, x), w(t, x) = e−i|x|24t u(t, x) and f (u) = λ|u|αu. Since T (t) is an
isometry on H1(RN), we only have to show that xT (−t)u(t)−−−−−→L2(RN)
t→±∞ xu±. We have X ֒→ L
α+2(RN),
thus u(t) − T (t)u±
Lα+2(RN)
−−−−−−→
t→±∞ 0 and so limt→±∞ku(t)kLα+2= limt→±∞kT (t)u±kLα+2= 0. Therefore, since
α > 4 N,
u ∈ Lq(R; W1,r(RN)), (5.1) v ∈ Lq(R; Lr(RN)), (5.2)
for every admissible pair (q, r) (see for example Cazenave [2], Theorem 7.5.3 for (5.1); following the proof of this theorem with v instead of u, yields (5.2)).
From (5.2) we have in particular, v ∈ L∞(R; L2(RN)) and so by Proposition3.1,1,
ku(t)kLr 6C|t|− 2
q, (5.3)
for every admissible pair (q, r) and for all t 6= 0. We have the following integral equation. For all t ∈ R,
u(t) = T (t)u±− i
±∞
Z
t
T (t − s)f(u(s))ds,
from which we deduce,
∀t ∈ R, T (t)(xT (−t)u(t) − xu±) = −i ±∞
Z
t
We also have (x + 2it∇)u(t, x) = 2itei|x|24t ∇w(t, x). Moreover, e−i |x|2
4t f (u(t, x)) = f (w(t, x)). Thus,
(x + 2it∇)f(u(t, x)) = 2itei|x|24t ∇f(w(t, x)), and so |(x + 2it∇)f(u(t, x))| = 2|t||∇f(w(t, x))|. Finally,
k(x+2it∇)f(u)kLr′ = 2|t|k∇f(w)kLr′6C|t|k|w|α∇wkLr′. From this inequality, by using the H¨older’s
inequality twice, we deduce that (note that |w| = |u|)
k(x + 2it∇)f(u)kLq′(I,Lr′)6Ckukα
L
qα
q−2(I,Lr−2rα )k(x + 2it∇)ukLq(I,Lr), (5.5)
for every admissible pair (q, r) and for any interval I ⊆ R. From (5.4), from the Strichartz’ estimates, from (5.5) and from (5.2), we have
kxT (−t)u(t) − xu+kL2= kT (t)(xT (−t)u(t) − xu+)kL2 6Ck(x + 2is∇)f(u)kLq′((t,∞);Lr′) 6Ckukα L qα q−2((t,∞);Lr−2rα )kvkLq(R;Lr) 6Ckukα L qα q−2((t,∞);Lr−2rα ),
for every admissible pair (q, r) and for all t > 0.
We use this inequality with the admissible pair (q, r) such that r = α + 2 and we apply (5.3). Then,
kxT (−t)u(t) − xu+kL26C Z ∞ t s−q−22αds q−2q 6Ct−2α−(q−2)q −−−→ 0,t→∞ since α > 4
N =⇒ 2α > q − 2. Hence the result. The case t < 0 follows with the same method.
Acknowledgments
I would like to express my great gratitude to Professor Thierry Cazenave, my thesis adviser, for having suggested me this work and for encouragement and helpful advices.
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